In the structure shown below members ABC, ADE,
and DB are assumed to be solid rigid members. Members ABC and ADE
are pinned to the wall at point A. Member ADE is supported by a
roller at point E. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the value of all the support forces acting on the
structure.
C. Determine the force (tension or compression) in member DB.
Unless otherwise indicated, all joints and support points are
assumed to be pinned or hinged joints.
Solution:
PARTS A & B
STEP 1:
Draw a free body
diagram showing and labeling all load forces and support
(reaction) forces, as well as any needed angles and dimensions.
STEP 2: Break any forces not already in x and y direction into
their x and y components.
STEP 3: Apply the equilibrium conditions.
Sum Fx = Ex + Ax = 0
Sum Fy = Ay - 12,000 lbs = 0
Sum TA = Ex(8 ft) - (12,000 lbs)(12 ft) = 0
Solving for the unknowns:
Ex= 18,000 lbs; Ax= -18,000 lbs; Ay
= 12,000 lbs
PART C - Now find internal force in member DB
STEP 1: 
Draw a free body
diagram of a member that DB acts on - member ABC.
STEP 2: Resolve all forces into x and y
components (see diagram)
STEP 3: Apply the equilibrium conditions.
Sum Fx = Acx + DB cos (33.7o) =
0
Sum Fy = Acy + DB sin (33.7o) -
12,000 lbs = 0
Sum TA = DB sin (33.7o)(6 ft) - (12,000
lbs)(12 ft) = 0
Solving for the unknowns:
DB = 43,300 lbs; Acx = -36,000 lbs; Acy = -12,000 lbs
These are external forces acting on member ABC.
The force in DB is 43,300 lbs (c).